\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{3^2\cdot2^2\cdot5^2\cdot7^{48}}\)

\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1+7\cdot4\right)}{3^2\cdot2^2\cdot5^2\cdot7^{48}}=\dfrac{5^{28}\cdot2^6\cdot12}{3^2}=\dfrac{5^{28}\cdot2^8}{3}\)

a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`

`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`

`=10(12x^2+25)-(120x^2+49)`

`=120x^2+250-120x^2-49`

`=201`

b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`

Lời giải:
a. 

$(2x+5)^3-(2x-5)^3-(120x^2+49)$

$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$

$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$

$=10(12x^2+25)-(120x^2+49)=250-49=201$

b.

$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$

$=7(1-10x)$

a) Ta có: \(\left(2x+5\right)^3-\left(2x-5\right)^3-\left(120x^2+49\right)\)

\(=8x^3+60x^2+150x+125-8x^3+60x^2-150x+125-120x^2-49\)

\(=201\)

b) Ta có: \(\left(4-5x\right)^2-\left(3+5x\right)^2\)

\(=\left(4-5x-3-5x\right)\left(4-5x+3+5x\right)\)

\(=7\left(-10x+1\right)\)

\(=-70x+7\)

\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{43}}\)

\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}\)

\(A=\frac{5^{30}.7^{48}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}=5.7^3.\left(1-7.2^2\right)=1715.\left(-27\right)=-46305\)

\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}\left(2^4\right)^2.7^{43}}=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}=\frac{7^{48}.5^{30}.2^8\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}\)

=\(7^5.5.\left(-27\right)=-2268945\)

A = 7⁵*5*-27

a, \(\left(5x-4\right)\left(5x+4\right)-\left(5x-4\right)^2=\left(25x^2-16\right)-\left(25x^2-40x+16\right)=40x-32\)

b,\(\left(5x+3\right)^2-\left(4x-1\right)^2-\left(9x^2+8\right)=\left(x+4\right)\left(9x-2\right)-\left(9x^2+8\right)\)

\(=9x^2+34x-8-\left(9x^2+8\right)=34x\)

c,\(2\left(x-5y\right)\left(x+5y\right)+\left(x+5y\right)^2+\left(x-5y\right)^2=\left(2x\right)^2=4x^2\)

\(\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)

\(=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)

\(=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

\(=\frac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)

\(=5.\left(1-7.4\right)\)

\(=5.\left(1-28\right)\)

\(=5.\left(-27\right)=-135\)

1. Tìm x, biết:

a) \(\frac{4}{x}=\frac{8}{6}\). Ta có: \(\frac{8}{6}=\frac{8:2}{6:2}=\frac{4}{3}\Rightarrow x=3\)

b) \(\frac{3}{x-5}=\frac{4}{x+2}\). Ta có: \(5-2=3\)

\(\Rightarrow x=\left(3.5\right)+\left(4.2\right)+3=15+8+3=26\)

c) \(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x=\left(-8\right):\left(-2\right)=4\)

2. Rút gọn

a) \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}\Leftrightarrow\frac{2^3.2^1.5^2.11.11.7}{2^3.5^2.5^1.7.7.11}\Leftrightarrow\frac{2^1.11}{5^1.7}=\frac{22}{35}\)

b) Tương tự

c) Tương tự

\(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{4.6}{8}=3\)

\(\frac{3}{x-5}=\frac{-4}{x+2}\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)

                                \(\Rightarrow3x+6=-4x+20\)

                                \(\Rightarrow3x+4x=20-6\)

                                \(\Rightarrow7x=14\)

                                \(\Rightarrow x=2\)

\(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x^2=\left(-8\right)\left(-2\right)=16\Rightarrow x=\pm4\)

\(\frac{4}{x}=\frac{8}{6}\)

\(\Rightarrow4\cdot6=x\cdot8\)

\(\Rightarrow x=\frac{4\cdot6}{8}\)

\(\Rightarrow x=3\)